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Expert-verifiedWhite light (consisting of wavelengths from to ) is normally incident on a grating. Show that, no matter what the value of the grating spacing , the second order and third order overlap.
It is proved no matter what the value of the grating spacing , the second order and third order overlap.
White light has wavelengths between
An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.
The angular distance of the order diffraction maxima is given by
…(i)
Here, is the wavelength of the incident light and is the separation between two lines in the grating.
From equation (i) the angular distance of second order maxima from is
From equation (i) the angular distance of third order maxima from is
Thus
Hence the angular distance of a part of the second order from white light is greater than the angular distance of a part of the third order. Thus the second and third order overlap.
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